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23-Chem-A3 Heat and Mass Transfer · May 2014

Question 4 of 7: Evaporation Rate from a Water Pool

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — May 2014 — 04-Chem-A3 Mass Transfer Operations. Three-hour, open-book exam; any non-communicating calculator permitted. Format: seven questions in three parts — answer one of Q1–Q2 (Part A), one of Q3–Q4 (Part B) and two of Q5–Q7 (Part C); four questions of equal value constitute a complete paper. All seven are solved below for completeness. Property data (diffusivities, vapour pressures, psychrometric enthalpies) are stated in each Given block; the psychrometric appendix supplied with the exam furnishes the saturated-air data used in Q6.

Reference texts: Geankoplis, Transport Processes and Separation Process Principles (4th ed., Prentice Hall) — molecular diffusion, convective mass transfer, absorption and humidification; Treybal, Mass-Transfer Operations (3rd ed., McGraw-Hill) — gas absorption, wetted-wall and cooling-tower design; Welty, Wicks, Wilson & Rorrer, Fundamentals of Momentum, Heat and Mass Transfer — boundary-layer and falling-sphere correlations; supporting property/psychrometric data from Perry's Chemical Engineers' Handbook (9th ed.).

Question 4: Evaporation Rate from a Water Pool (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. External laminar flow over a flat liquid surface; the mean flat-plate Sherwood correlation gives $k_c$, and the surface vapour pressure sets the interface concentration.

QuantityValue
Air state300 K, 1.0 atm, $u=0.5$ m/s
Pool$L=10$ m (flow direction) × 4 m wide
Water surface287 K, $p^{vap}=1620$ Pa
$\rho_{air}$, $\mu_{air}$1.1769 kg/m³, $1.8464\times10^{-5}$ Pa·s
$D_{H_2O\text{-}air}$$2.45\times10^{-5}$ m²/s

Find. The total evaporation rate from the pool.

water at 287 Kair 300 K, u = 0.5 m/s →water vapour transferpool length L = 10 m (× 4 m wide)
Figure 4 — Dry air blows along a 10 m × 4 m pool of water held at 287 K; a laminar concentration boundary layer grows over the pool and water evaporates into the moving air.

Approach. Check the flow regime with $Re_L$, apply the mean laminar flat-plate correlation for $\overline{Sh}$, get $k_c$, and multiply the surface–bulk concentration difference (surface value at the water temperature) by the pool area.

  1. Reynolds number over the pool. $Re_L = \dfrac{\rho u L}{\mu} = \dfrac{(1.1769)(0.5)(10)}{1.8464\times10^{-5}} = 3.19\times10^{5} < 5\times10^{5}$, so the boundary layer is laminar over the whole pool.
  2. Schmidt number and mean Sherwood number. $Sc = \dfrac{\mu}{\rho D} = 0.640$; $\ \overline{Sh} = 0.664\,Re_L^{1/2}Sc^{1/3} = 0.664(564.5)(0.862) = 323.$
  3. Mean mass-transfer coefficient. $$\overline{k_c} = \frac{\overline{Sh}\,D}{L} = \frac{323(2.45\times10^{-5})}{10} = \boxed{7.9\times10^{-4}\ \text{m/s}}.$$
  4. Surface vapour concentration. Evaluated at the water-surface temperature (287 K): $c_{As} = p^{vap}/RT = 1620/(8.314\times287) = 0.679\ \text{mol/m}^3$; the dry bulk air gives $c_{A\infty}=0$.
  5. Evaporation rate over the pool. With area $A=10\times4=40\ \text{m}^2$, $$\dot m = \overline{k_c}\,c_{As}\,A\,M_{H_2O} = (7.9\times10^{-4})(0.679)(40)(0.018) = \boxed{3.9\times10^{-4}\ \text{kg/s}\ (1.39\ \text{kg/h})}.$$
QuantityResult
$Re_L$ (laminar)$3.19\times10^{5}$
$\overline{k_c}$$7.9\times10^{-4}$ m/s
Surface conc. $c_{As}$0.679 mol/m³
Evaporation rate≈ 3.9×10⁻⁴ kg/s (0.39 g/s, 1.39 kg/h)
Check
Two temperatures matter: the transport coefficient uses the flowing-air properties (300 K air), but the surface concentration must use the water-surface temperature (287 K), where the vapour pressure is evaluated. Mixing them up is the classic error in this problem.